Q55P - Mathematical Methods in Physical Sciences

Question

Find the inverse Laplace transform of the following functions using (7.16) $\frac{{\mathbf{p}}^{\mathbf{3}}}{{\mathbf{p}}^{\mathbf{4}}\mathbf{+}\mathbf{4}}$.

Step-by-Step Solution

Answer

The residues at poles, $\mathit{f}\left(t\right)\mathbf{=}\left(cosht\right)\left(cost\right)$

Step 1: Consider the real and imaginary part.

Apply complex number form,

 $\mathit{z}\mathbf{ }\mathbf{=}\mathbf{ }\mathit{x}\mathbf{ }\mathbf{+}\mathbf{ }\mathit{i}\mathit{y}$

 

The real part is x  and the imaginary part of the complex number is y.

The polar form is called the  $\mathit{r}\mathbf{\times }{\mathit{e}}^{\mathbf{i}\mathbf{\theta }}$ representation of a complex number in the form.

Step 2: Determine the pole

Consider the function of the given equation.

 

  $\mathit{F}\left(p\right)\mathbf{=}\frac{{\mathbf{p}}^{\mathbf{3}}}{{\mathbf{p}}^{\mathbf{4}}\mathbf{+}\mathbf{4}}$

 

To find the pole of $\mathit{F}\left(z\right){\mathit{e}}^{\mathbf{z}\mathbf{t}}$  and factor the denominator to obtain.

 

  $\begin{array}{rcl}\mathit{F}\left(z\right){\mathit{e}}^{\mathbf{z}\mathbf{t}}& \mathbf{=}& \frac{{\mathbf{z}}^{\mathbf{3}}{\mathbf{e}}^{\mathbf{z}\mathbf{t}}}{\left(z^2-2z+2\right)\left(z^2+2z+2\right)}\\ & \mathbf{=}& \frac{{\mathbf{z}}^{\mathbf{3}}{\mathbf{e}}^{\mathbf{2}\mathbf{z}}}{\left(z-\alpha \right)\left(z-\beta \right)\left(z+\alpha \right)\left(z+\beta \right)}\end{array}$

 

Consider the values of the variables for $\mathit{\alpha }\mathbf{,}\mathbf{ }\mathit{\beta }\mathbf{,}\mathbf{ }\mathbf{-}\mathit{\alpha }$  and $\mathbf{-}\mathit{\beta }$ of the function.

  $\begin{array}{rcl}\mathit{\alpha }& \mathbf{=}& \mathbf{1}\mathbf{+}\mathit{i}\\ \mathit{\beta }& \mathbf{=}& \mathbf{1}\mathbf{-}\mathit{i}\\ \mathbf{-}\mathit{\alpha }& \mathbf{=}& \mathbf{-}\mathbf{1}\mathbf{-}\mathit{i}\\ \mathbf{-}\mathit{\beta }& \mathbf{=}& \mathbf{-}\mathbf{1}\mathbf{+}\mathit{i}\end{array}$

Step 3: Determine the residues at four poles

So now we find the residues at the four poles.

Residue at,  $\mathit{z}\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{1}\mathbf{ }\mathbf{+}\mathbf{ }\mathit{i}$

 $\begin{array}{rcl}\mathbf{Re}\mathbf{s}\left(z=1+i\right)& \mathbf{=}& \frac{{\left(1+i\right)}^{\mathbf{3}}{\mathbf{e}}^{\left(1+i\right)\mathbf{t}}}{\left(1+i-1+i\right)\left(1+i+1+i\right)\left(1+i+1-i\right)}\\ & \mathbf{=}& \frac{{\left(1+i\right)}^{\mathbf{3}}{\mathbf{e}}^{\mathbf{t}}\mathbf{\times }{\mathbf{e}}^{\mathbf{it}}}{\mathbf{2}\mathbf{i}\mathbf{\times }\mathbf{2}\mathbf{\times }\mathbf{2}\left(1+i\right)}\\ & \mathbf{=}& \frac{{\mathbf{e}}^{\left(1+i\right)\mathbf{t}}}{\mathbf{4}}\end{array}$

  

 

Residue at, $\mathit{z}\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{1}\mathbf{ }\mathbf{-}\mathbf{ }\mathit{i}$ 

 $\begin{array}{rcl}\mathbf{Re}\mathit{s}\left(z=1-i\right)& \mathbf{=}& \frac{{\left(1-i\right)}^{\mathbf{3}}{\mathbf{e}}^{\left(1-i\right)\mathbf{t}}}{\left(1-i-1-i\right)\left(1-i+1+i\right)\left(1-i+1-i\right)}\\ & \mathbf{=}& \frac{{\left(1-i\right)}^{\mathbf{3}}{\mathbf{e}}^{\left(1-i\right)\mathbf{t}}}{\left(-2i\right)\mathbf{\times }\mathbf{2}\mathbf{\times }\mathbf{2}\mathbf{\times }\left(1-i\right)}\\ & \mathbf{=}& \frac{{\mathbf{e}}^{\left(1-i\right)\mathbf{t}}}{\mathbf{4}}\end{array}$

 

 

Residue at, $\mathit{z}\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{ }\mathbf{-}\mathbf{ }\mathbf{1}\mathbf{ }\mathbf{-}\mathbf{ }\mathit{i}$

 $\begin{array}{rcl}\mathbf{Re}\mathit{s}\left(z=-1-i\right)& \mathbf{=}& \frac{{\left(-1-i\right)}^{\mathbf{3}}\mathbf{\times }{\mathbf{e}}^{{\left(-1-i\right)}^{\mathbf{t}}}}{\left(-1-i-1-i\right)\left(-1-i-1+i\right)\left(-1-i+1-i\right)}\\ & \mathbf{=}& \frac{{\left(-1-i\right)}^{\mathbf{3}}{\mathbf{e}}^{\left(-1-i\right)\mathbf{t}}}{\mathbf{2}\left(-1-i\right)\left(-2\right)\left(-2i\right)}\\ & \mathbf{=}& \frac{{\mathbf{e}}^{\mathbf{-}\left(1+i\right)\mathbf{t}}}{\mathbf{4}}\end{array}$

 

 

Residue at,  $\mathit{z}\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{ }\mathbf{-}\mathbf{ }\mathbf{1}\mathbf{ }\mathbf{+}\mathbf{ }\mathit{i}$

$\mathbf{Re}\mathit{s}\left(z=-1-i\right)\mathbf{=}\frac{{\mathbf{e}}^{\left(-1+i\right)\mathbf{t}}}{\mathbf{4}}$

 

Step 4: To find the value of f ( t )

Sum of all residues of $\mathit{F}\left(z\right){\mathit{e}}^{\mathbf{2}\mathbf{t}}$ at poles is:

$\begin{array}{rcl}\mathit{f}\left(t\right)& \mathbf{=}& \frac{{\mathbf{e}}^{\mathbf{t}}\mathbf{\times }{\mathbf{e}}^{\mathbf{i}\mathbf{t}}}{\mathbf{4}}\mathbf{+}\frac{{\mathbf{e}}^{\mathbf{t}}\mathbf{\times }{\mathbf{e}}^{\mathbf{-}\mathbf{i}\mathbf{t}}}{\mathbf{4}}\mathbf{+}\frac{{\mathbf{e}}^{\mathbf{-}\mathbf{t}}\mathbf{\times }{\mathbf{e}}^{\mathbf{-}\mathbf{i}\mathbf{t}}}{\mathbf{4}}\mathbf{+}\frac{{\mathbf{e}}^{\mathbf{-}\mathbf{t}}\mathbf{\times }{\mathbf{e}}^{\mathbf{i}\mathbf{t}}}{\mathbf{4}}\\ & \mathbf{=}& \frac{{\mathbf{e}}^{\mathbf{t}}}{\mathbf{2}}\left[\frac{e^{it}+e^{-it}}{2}\right]\mathbf{+}\frac{{\mathbf{e}}^{\mathbf{-}\mathbf{t}}}{\mathbf{2}}\left[\frac{e^{it}+e^{-it}}{2}\right]\\ & \mathbf{=}& \left(\frac{e^t+e^{-t}}{2}\right)\mathit{c}\mathit{o}\mathit{s}\mathit{t}\\ & \mathbf{=}& \left(cosht\right)\left(cost\right)\end{array}$

Hence,  $\mathit{f}\left(t\right)\mathbf{=}\left(cosht\right)\left(cost\right)$.